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_bTZbn7M2Hc | So this becomes the sum
over k different from n, |
_bTZbn7M2Hc | because for k equal to n, these
are orthogonal of k0 k0 n1. |
_bTZbn7M2Hc | And that's what we
calculated here. |
_bTZbn7M2Hc | So what did we get? |
_bTZbn7M2Hc | Therefore, the state n1, I can
substitute what we had there. |
_bTZbn7M2Hc | It's the sum from k different
from n of k0 delta Hnk over Ek0 |
_bTZbn7M2Hc | minus En0. |
_bTZbn7M2Hc | That's n1. |
_bTZbn7M2Hc | I should have a minus sign. |
_bTZbn7M2Hc | The minus sign is
there at the state n1. |
_bTZbn7M2Hc | So the state n1 is a
complicated correction. |
_bTZbn7M2Hc | It gets a little component
from every other state |
_bTZbn7M2Hc | of the spectrum. |
_bTZbn7M2Hc | And the coefficient depends
on the matrix element |
_bTZbn7M2Hc | of your state with the state
you're contributing with. |
_bTZbn7M2Hc | So you have the state n and
all the other states here. |
_bTZbn7M2Hc | The amount of this state k
that enters into the correction |
_bTZbn7M2Hc | is proportional to the matrix
element between n and k. |
_bTZbn7M2Hc | If the matrix element is 0, that
state does not contribute here. |
_bTZbn7M2Hc | And then there is the
energy denominator as well. |
_bTZbn7M2Hc | So we're getting to the
end of this calculation. |
_bTZbn7M2Hc | There's one more thing one
can do, which is to find-- |
_bTZbn7M2Hc | so I'm starting to
wrap up this, but still |
_bTZbn7M2Hc | an important step
what we have to do. |
_bTZbn7M2Hc | I'll get the second
order energy correction. |
_bTZbn7M2Hc | What is our second
order energy correction? |
_bTZbn7M2Hc | Our second order
energy correction |
_bTZbn7M2Hc | can be found from the formula
on that blackboard, En2. |
_bTZbn7M2Hc | We already found the first
order energy correction, |
_bTZbn7M2Hc | which I happened to have
erased it right now. |
_bTZbn7M2Hc | It was there. |
_bTZbn7M2Hc | En2 is obtained by doing
n0 delta H times n1, which |
_bTZbn7M2Hc | we already know. |
_bTZbn7M2Hc | So I must do n0 delta H on that. |
_bTZbn7M2Hc | So look what you get. |
_bTZbn7M2Hc | You get minus the sum
over k different from n. |
_bTZbn7M2Hc | Think of putting the n0 and the
delta H, they're all together. |
_bTZbn7M2Hc | It's a [INAUDIBLE] so far. |
_bTZbn7M2Hc | It's a delta H and n0. |
_bTZbn7M2Hc | IT should go into n1. |
_bTZbn7M2Hc | But the only state in n1 is k0. |
_bTZbn7M2Hc | So here we have k0. |
_bTZbn7M2Hc | And then we have delta
Hnk over Ek0 minus En0. |
_bTZbn7M2Hc | OK. |
_bTZbn7M2Hc | A little bit of work. |
_bTZbn7M2Hc | So what is this? |
_bTZbn7M2Hc | This is another matrix element. |
_bTZbn7M2Hc | This is the matrix-- |
_bTZbn7M2Hc | OK. |
_bTZbn7M2Hc | I'm sorry. |
_bTZbn7M2Hc | Here do I have a mistake? |
_bTZbn7M2Hc | Oh, yes, I have kn. |
_bTZbn7M2Hc | I copied it wrong. |
_bTZbn7M2Hc | It's kn. |
_bTZbn7M2Hc | Yes. |
_bTZbn7M2Hc | Yes. |
_bTZbn7M2Hc | So here I have delta Hnk. |
_bTZbn7M2Hc | but delta Hnk is this. |
_bTZbn7M2Hc | If you complex conjugate-- |
_bTZbn7M2Hc | if you complex
conjugate delta Hkn, |
_bTZbn7M2Hc | complex conjugate is k
delta H n complex conjugate, |
_bTZbn7M2Hc | which changes the order. |
_bTZbn7M2Hc | n delta H, which
is her mission k. |
_bTZbn7M2Hc | And that's delta Hnk. |
_bTZbn7M2Hc | So delta Hnk is equal
to delta Hkn star. |
_bTZbn7M2Hc | And therefore the second
order energy correction |
_bTZbn7M2Hc | has a nice formula. |
_bTZbn7M2Hc | En2 is equal to minus the
sum over k different from n. |
_bTZbn7M2Hc | Delta Hnk, which is the
star of that times this one, |
_bTZbn7M2Hc | so you get delta Hnk absolute
value squared divided by Ekn. |
_bTZbn7M2Hc | Ek0 minus En0. |
_bTZbn7M2Hc | So we've done a lot of work. |
_bTZbn7M2Hc | We've written the perturbation. |
_bTZbn7M2Hc | Here is the answer. |
_bTZbn7M2Hc | So far we have n of lambda
equal n0 plus lambda n1. |
_bTZbn7M2Hc | n1 has been calculated. |
_bTZbn7M2Hc | Energy is En0 plus lambda En1. |
_bTZbn7M2Hc | That was calculated
what was just |
_bTZbn7M2Hc | delta H in this state
plus lambda squared |
_bTZbn7M2Hc | En2, which we have calculated. |
_bTZbn7M2Hc | So this is as far as we will do
for nondegenerate perturbation |
_bTZbn7M2Hc | theory. |
_bTZbn7M2Hc | But we have found rather
interesting formulas. |
_bTZbn7M2Hc | And we're going to spend
half of its lecture trying |
_bTZbn7M2Hc | to understand them better. |
rLlZpnT02ZU | The following content is
provided under a Creative |
rLlZpnT02ZU | Commons license. |
rLlZpnT02ZU | Your support will help
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rLlZpnT02ZU | continue to offer high quality
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rLlZpnT02ZU | To make a donation or to
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rLlZpnT02ZU | PHILIPPE RIGOLLET: --124. |
rLlZpnT02ZU | If I were to repeat
this 1,000 times, |
rLlZpnT02ZU | so every one of
those 1,000 times |
rLlZpnT02ZU | they collect 124
data points and then |
rLlZpnT02ZU | I'd do it again and
do it again and again, |
rLlZpnT02ZU | then in average, the
number I should get |
rLlZpnT02ZU | should be close to the true
parameter that I'm looking for. |
rLlZpnT02ZU | The fluctuations that
are due to the fact |
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